A scientist has devised a new method of isolating individual particles. He claims that this method enables him to detect simultaneously the position of a particle along an axis with a standard deviation of 0.12 nm and its momentum component along this axis with a standard deviation of 3.0x10^-25 kg-m/s. Use the Heisenberg uncertainty principle to evaluate the validity of this claim.

Video transcript

Hello everyone. Let's go through this problem. You are reviewing a scientific paper presenting a new method to measure simultaneously the position and the velocity of electrons and atoms. The authors estimate the uncertainty on the position of the electrons at 10 to the power of a negative nine m with delta X is 10 to the power of negative nine m. And the uncertainty on the velocity at 3.3 multiplied by 10 to the power of four m per second. So delta V the uncertainty and the velocity is 3.3 multiplied by 10 to the power of four m per second. Will you agree on publishing this paper? And we have four multiple choice options. Option. A yes. The uncertainty on the position of the electron is very low. Option B yes, the results are in accordance with all the principles of quantum mechanics. Option C no, the uncertainty on the velocity of the electrons is very high and option D know the results violate Heisenberg's principle. Now, one way we can check the validity of results in quantum mechanics is to check to see if the results agree with the Heisenberg uncertainty principle. And recall that the Heisenberg uncertainty principle states that the uncertainty in the position of a particle multiplied by the uncertainty in the momentum of the particle must be greater than or equal to the plank constant divided by four pi. And recall that the plant constant is equal to 6.626 multiplied by 10 to the power of negative 34 dual seconds. Now, the problem doesn't tell us what the uncertainty in momentum is, but recall that momentum is equal to the mass multiplied by the velocity. So the mass of the electron multiplied by the uncertainty of the velocity which we are given can substitute in for the uncertainty in the momentum. And since we're dealing with an electron recall that the mass of an electron is equal to 9.11 multiplied by 10 to the power of negative 31 kg. So now to test this result, let's plug in the values we have. So for delta X, we're plugging in 10 to the power of negative nine m. For M we're plugging in 9.11 multiplied by 10 to the power of negative 31 kg. For delta V, we're plugging in 3.3 multiplied by 10 to the power of four m per second. And this is greater than or equal to the plant constant 6.626 multiplied by 10 to the power of negative 34 dual seconds. All divided by four pi So now let's plug both sides of this into a calculator to expand it out and get a single numerical result. So we find that the left hand side of the equation is equal to 3. multiplied by 10 to the power of negative 35 dual seconds. And the right hand side is equal to about 5. multiplied by 10 to the power of a negative 35 dual seconds. And just from looking at this, we can see there is a big error because the left hand side of this equality is clearly smaller than the right hand side. So Heisenberg's uncertainty principle is violated. And if we look at our multiple choice options, we can see that this is basically what option D says that no, the results violate Heisenberg's uncertainty principle. So the results here are not valid. And so option D is the answer to the problem and that's it. I hope this video helped you out. If you'd like more practice, please check out some of our other videos which will hopefully give you more experience with these types of problems, but that's all for now. And I hope you all have a lovely day. Bye bye.